SME
Chemical Engineering Thermodynamics · Problem type
Equilibrium constant at another temperature
Use this for “The water–gas shift has ΔH° = −41.2 kJ/mol and K = 1 × 10⁵ at 298.15 K. Find K at 500 K.”
Chart
ln K₁ = 11.1x₁ − 23.9
Slope m = 11.1: up 11.1 for every 1 across
Intercept b = −23.9: it crosses the y-axis at (0, −23.9)
ln K₁ = 11.1 × 3.35 − 23.9 = 13.3 when x₁ = 3.35
This is an example. Type your own values to start a new problem.
ln
K₂
K₁
=
−(
ΔH° (kJ/mol)
÷
R)
×
1
T₂ (K)
−
1
T₁ (K)
Formulas
ln(
K₂
K₁
)
=
−ΔH°
×
1,000
8.314
×
(
1
T₂
−
1
T₁
)
ln(
3.10 ×
10
-4
5.81 ×
10
5
)
=
−(−92.2)
×
1,000
8.314
×
(
1
700
−
1
298.15
)
Assumptions
•
ΔH° stays constant between T₁ and T₂, so ln K against 1/T is a straight line of slope −ΔH° ÷ R.
•
R = 8.314 J/(mol·K); ΔH° in kJ/mol is multiplied by 1000. Temperatures are in kelvins.
•
Exothermic reactions (ΔH° < 0) lose K as they get hotter; endothermic ones gain K.
•
The graph plots ln K against x = 1000/T, so its slope reads −ΔH° ÷ R with ΔH° in kJ/mol.
Step-by-step
WE KNOW
ΔH° = −92.2 kJ/mol, K₁ = 5.81 × 10⁵, T₁ = 298.15 K, T₂ = 700 K
FIND
equilibrium constant at T₂ (K₂)
Step 1 · Find equilibrium constant at T₂ (K₂)
Use ln(K₂ ÷ K₁) = −ΔH° × 1,000 ÷ 8.314 × (1 ÷ T₂ − 1 ÷ T₁)
Work
out
how
much
ln
K
changes
from
T₁
to
T₂
,
raise
e
to
it,
and
scale
K₁
by
that.
K₂
=
K₁
×
e^(−ΔH°
×
1,000
÷
8.314
×
(
1
T₂
−
1
T₁
))
K₂
=
5.81 ×
10
5
×
e^(−(−92.2)
×
1,000
÷
8.314
×
(
1
700
−
1
298.15
))
K₂
=
(5.81
×
10
5
)
×
e
−(−92.2) × 1,000 ÷ 8.314 × (0.00142857 − 0.00335402)
K₂
=
(5.81
×
10
5
)
×
e
−(−92.2) × 1,000 ÷ 8.314 × (−0.00192545)
K₂
=
(5.81
×
10
5
)
×
e
92.2 × 1,000 ÷ 8.314 × (−0.00192545)
K₂
=
(5.81
×
10
5
)
×
e
92,200 ÷ 8.314 × (−0.00192545)
K₂
=
(5.81
×
10
5
)
×
e
11,089.7 × (−0.00192545)
K₂
=
(5.81
×
10
5
)
×
e
−21.3527
K₂
=
(5.81
×
10
5
)
×
(5.32893
×
10
-10
)
K₂
=
3.10 ×
10
-4
Check
Put the numbers back in. Both sides should match.
ln(
3.10 ×
10
-4
5.81 ×
10
5
)
=
−(−92.2)
×
1,000
8.314
×
(
1
700
−
1
298.15
)
✓
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